Question

A system being studied in a 3.0-liter reactor was charged with 0.150 moles of H2, 0.150 moles of X2, and 0.600 moles of HX. A catalyst was introduced using a remote unit. Is the reaction at equilibrium

Answer 1

The reaction is not in equilibrium and it proceed in reverse direction to reach the equilibruim condition

Step-by-step explanation:

We have

H₂(g) + x₂(g) ⇆ 2HX (g)

Kc = 2.4 x 10 ⁻³ at 25°C

`K_c = ([HX]^2)/([H_2][X_2])`

Now we have 3L reactor and the number of mole of species are 0.150 mole H₂, 0.150 mole X₂ and 0.600 mole HX

Therefore,

`[H_2] = (0.150)/(3) \\= 0.05M`

`[X_2] = (0.150)/(3) \\= 0.05M`

`[HX] = (0.600)/(3) \\= 0.2M`

Therefore,

`Q_c = ([HX]^2)/([H_2][X_2])`

`Q_c = ([0.200]^2)/([0.05][0.05])\\\\=16`

Now, we calculate the free energy change for the reaction

ΔrG = RT In Q/K

`= (8.314) * (298) * In ((16)/(2.4* 10^-^3) )\\\\=+21814.7 J/mol`

As free energy is +ve so the reaction is not in equilibruim and proceed in reverse reaction to reach tje equilibruim condition